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  • 07-10-2020
  • Mathematics
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What are the roots of f(x)=(x-6)^2(x+2)^2

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11qq11
11qq11 11qq11
  • 07-10-2020

f(x)=0

⇒(x−6)^2. (x+2) ^2 =0

⇒(x−6)^2=0, (x+2)^2=0

⇒x=6,6 ⇒x=−2,−2.

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